C:UsersBubakazouba p1 2 a 1 - solution banks sol FP1/CH2.… · Solutionbank FP1 Edexcel AS and A Level Modular Mathematics Numerical solutions of equations Exercise A, Question 2

Solutionbank FP1 Edexcel AS and A Level Modular Mathematics Numerical solutions of equations Exercise A, Question 1

© Pearson Education Ltd 2008

Question:

Use interval bisection to find the positive square root of , correct to one decimal place. x2 − 7 = 0

Solution:

So roots lies between 2 and 3 as and Using table method.

Hence x2 − 7 = 0 when x = 2.6 to 1decimal place

x2 − 7 = 0

f(2) = −3 f(2) = +

a f(a) b f(b) a + b

2

f(a + b)

2

2 −3 3 +2 2.5 −0.752.5 −0.75 3 +2 2.75 0.56252.5 −0.75 2.75 0.5625 2.625 −0.109375

2.625 −0.109375 2.75 0.5625 2.6875 0.22265622.625 −0.109375 2.6875 0.2226562 2.65625 0.0556642.625 −0.109375 2.65625 0.055664 2.640625 −0.0270996

Page 1 of 1Heinemann Solutionbank: Further Pure FP1

3/18/2013file://C:\Users\Buba\kaz\ouba\fp1_2_a_1.html

http://www.kvisoft.com/pdf-merger/



Question:

a Show that one root of the equation lies in the interval [2, 3].

b Use interval bisection to find the root correct to two decimal places.

x3− 7x + 2 = 0

Solution:

a

Hence change of sign, implies roots between 2 and 3.

b Using table method.

Hence x = 2.49 to 2 decimal places.

f(2) = 8 − 14+ 2 = −4 f(x) = x3− 7x + 2

f(3) = 27− 21+ 2 = +8

a f(a) b f(b) a + b

2

f(a + b)

2

2 −4 3 +8 2.5 0.1252 −4 2.5 0.125 2.25 −2.359375

2.25 −2.359375 2.5 0.125 2.375 −1.22851562.375 −1.2285156 2.5 0.125 2.4375 −0.58032222.4375 −0.5803222 2.5 0.125 2.46875 −0.23489382.46875 −0.2348938 2.5 0.125 2.484375 −0.05676652.484375 −0.0567665 2.5 0.125 2.4921875 0.03366042.484375 −0.0567665 2.4921875 0.0336604 2.4882813 −0.0116673





Question:

a Show that the largest positive root of the equation lies in the interval [2, 3].

b Use interval bisection to find this root correct to one decimal place.

0 = x3 + 2x2 − 8x − 3

Solution:

a

Change of sign implies root in interval [2,3]

b

Hence solution = 2.2 to 1decimal place

f(2) = 8 + 8 − 16− 3 = −3 f(x) = x3+ 2x2 − 8x − 3

f(3) = 27+ 18− 24− 3 = 18

a f(a) b f(b) a + b

2f( a + b

2 )2 −3 3 18 2.5 5.1252 −3 2.5 5.125 2.25 0.515622 −3 2.25 0.515625 2.125 −1.37304

2.125 −1.3730469 2.25 0.515625 2.1875 −0.46215





Question:

a Show that the equation has one root which lies in the interval [0.5, 0.8].

b Use interval bisection four times to find this root. Give your answer correct to one decimal place.

f(x) = 1 − 2sinx

Solution:

a

Change of sign implies root between 0.5 and 0.8

b

0.5 to 1 decimal place.

f(0.5) = +0.0411489

f(0.8) = −0.4347121

a f(a) b f(b) a + b

2

f(a + b)

2

0.5 0.0411489 0.8 −0.4347121 0.65 −0.21037280.5 0.0411489 0.65 −0.2103728 0.575 −0.08766950.5 0.0411489 0.575 −0.0876696 0.5375 −0.0239802





Question:

a Show that the equation , has a root in the interval [1, 2].

b Obtain the root, using interval bisection two times. Give your answer to two significant figures.

0 = x

2− 1

x, x > 0

Solution:

a

Change of sign implies root between interval [1,2]

b

Hence to 2 significant figures

f(1) = −0.5 p = 1

2+ x − 1

x

f(2) = +0.5

a f(a) b f(b) a + b

2f( a + b

2 )1 −0.5 2 +0.5 1.5 0.08331 −0.5 1.5 0.083 1.25 −0.175

1.25 −0.175 1.5 0.083 1.375 −0.03977271.375 −0.0397727 1.5 0.083 1.4375 0.0230978

x = 1 . 4





Question:

The equation has a root between and . Starting with the interval [2, 3] use interval bisection three times to give an approximation to this root.

f(x) = 6x − 3x

f(x) = 0 x = 2 x = 3

Solution:

2.4 correct to 1 decimal place.

a f(a) b f(b) a + b

2f( a + b

2 )2 3 3 −9 2.5 −0.5884572 3 2.5 −0.5884572 2.25 1.65533

2.25 1.6553339 2.5 −0.5884572 2.375 0.661762.375 0.6617671 2.5 −0.5884572 2.4375 0.07082.4375 0.0709769 2.5 −0.5844572 2.46875 −0.24982.4375 0.0709769 2.46875 −0.2498625 2.453125 −0.087262.4375 0.0709769 2.453125 −0.0872613 2.4453125 −0.0076



Solutionbank FP1 Edexcel AS and A Level Modular Mathematics Numerical solutions of equations Exercise B, Question 1

Question:

a Show that a root of the equation lies in the interval [2, 3].

b Find this root using linear interpolation correct to one decimal place.

x3− 3x − 5 = 0

Solution:

a

Change of size therefore root in interval [2, 3]

b Using linear interpolation and similar triangle taking x1 as the first root.

so

Using interval (2, 2.8125)



Using interval (2.112, 2.375)


f(2) = 8 − 6 − 5 = −3 f(x) = x3− 3x − 5

f(3) = 27− 9 − 5 = +13

3 − x1x1− 2

= 3

13x = af(b) − bf(a)

f(b) − f(a)

13(3− x1) = 3(x1− 2)39− 13x1 = 3x1− 6

16x1 = 45

x1 = 2.8125 f(x1) = 8.8098

2.8125− x2x2 − 2

= 3

8.8098

x2 = 2.606 f(x2) = 4.880

2.606− x3x3 − 2

= 34.880

x2 = 2.375 f(x2) = 1.276

2.375− x4x4 − 2

= 3

1.276

x2 = 2.112 f(x4) = −1.915

2.375− x5x5 − 2.112

= 1.915

1.276

= 2.218 f(x5) = −0.736


3/18/2013file://C:\Users\Buba\kaz\ouba\fp1_2_b_1.html



2.3 to 1 decimal place.

2.375− x6x6 − 2.218

= 0.736

1.276

= 2.318 f(x6) = 0.494

2.318− x7x7 − 2.218

= 0.736

0.494

= 2.25 f(x7) = −0.229





Question:

a Show that a root of the equation has a root between and .

b Find this root using linear interpolation correct to one decimal place.

5x3 − 8x2 + 1 = 0 x = 1 x = 2

Solution:

a

Therefore root in interval [1, 2] as sign change.

b Using linear interpolation.




Use interval (1.447, 1.570)

Ans 1.5 correct to 1 decimal place.

f(1) = 5 − 8 + 1 = −2 f(x) = 5x3− 8x2 + 1

f(2) = 40− 32+ 1 = +9

2 − x1x1− 1

= 2

9

x1 = 1.818 f(x1) = 4.612.

1.818− x2x2 − 1

= 2

4.612

x2 = 1.570 f(x2) = 0.647

1.570− x3x3− 1

= 2

0.647

x3 = 1.139 f(x3) = −1.984

1.570− x4x4 − 1.139

= 1.984

0.647

x4 = 1.447 f(x4) = −0.590

1.570− x5x5 − 1.447

= 0.590

0.647

= 1.511 f(x5) = −0.0005.





Question:

a Show that a root of the equation lies in the interval [3, 4].

b Use linear interpolation to find this root correct to one decimal place.

3

x+ 3 = x

Solution:

a

Hence root as sign change in interval [3, 4]

b Using linear interpolation

Using interval [3, 3.8]

Ans = 3.8 to 1decimal place

f(3) = 1 f(x) = 3

x+ 3 − x

f(4) = −0.25

4 − x1x1− 3

= 0.25

1

x1 = 3.8 f(x1) = −0.011

3.8− x2x2 − 3

= 0.0111

1

x2 = 3.791 f(x2) = −0.0004579





Question:

a Show that a root of the equation lies in the interval [1, 1.5].

b Find this root using linear interpolation correct to two decimal places.

2x cosx − 1 = 0

Solution:

a

Hence root between (1, 1.5) as sign change


Use interval [1, 1.28]




Root 1.10 to 2 decimal places.

f(1) = 0.0806

f(1.5) = −0.788

1.5− x1x1− 1

= 0.788

1

x1 = 1.280 f(1.280)= −0.265

1.28− x2x2 − 1

= 0.265

1

x2 = 1.221 f(1.221)= −0.164

1.221− x2x3− 1

= 0.164

1

x3 = 1.190 f(1.190)= −0.115

1.190− x4x4 − 1

= 0.115

1

x4 = 1.170 f(1.170)= 0.088

1.170− x5x5 − 1

= 0.088

1

x5 = 1.156 f(1.156)= −0.068





Question:

a Show that the largest possible root of the equation lies in the interval [2, 3].

b Find this root correct to one decimal place using interval interpolation.

x3− 2x2 − 3 = 0

Solution:

a

Hence root lies in interval [2, 3] and .


Hence root = 2.5 to 1 d.p

f(2) = 8 − 8 − 3 = −3 f(x) = x3− 2x2 − 3

f(3) = 27− 18− 3 = 6

∀ x ∈ x ≥ 3f(x) < 0

3 − x1x1− 2

= 6

3

x1 = 2.333 f(x1) = −1.185

3 − x2x2 − 2.333

= 6

1.185

x2 = 2.443 f(x2) = −0.356

3 − x3x3 − 2.443

= 6

0.356

x3 = 2.474 f(x3) = −0.095

3 − x4x4 − 2.474

= 6

0.095

x4 = 2.482





Question:

The equation has a root in the interval [3, 4].

Using this interval find an approximation to x.

f(x) = 2x − 3x − 1

f(x) = 0

Solution:

Let root be α

f(3) = −2f(4) = 3

4 − αα − 3

= 3

2α = 3.4 is the approximation.



Solutionbank FP1 Edexcel AS and A Level Modular Mathematics Numerical solutions of equations Exercise C, Question 1


Question:

Show that the equation has a root between 1 and 2. Find the root correct to two decimal places using the Newton–Raphson process.

x3− 2x − 1 = 0

Solution:

Hence root in interval [1,2] as sign change

Let x0 = 2.

f(1) = −2 f(x) = x3 − 2x − 1f(2) = 3 f(2) = 3 is correct

f(x) = x3− 2x − 1

f ′(x) = 3x2 − 2

Then x1 = x0 −f(x0)

f ′(x0)

x1 = 2 − 310

x1 = 1.7

x2 = x1−f(x1)

f ′(x1)

x2 = 1.88− 1.8858.6032

= 1.661

x3 = x2 − f(x2)

f ′(x2)

x3 = 1.661− 0.2597

6.2767= 1.6120

x4 = 1.620− f(1.620)

f ′(1.620)

x4 = 1.62− 0.0115

5.8732= 1.618

Solution = 1.62 to 2 decimal places


3/18/2013file://C:\Users\Buba\kaz\ouba\fp1_2_c_1.html



Question:

Use the Newton–Raphson process to find the positive root of the equation correct to two decimal places.

x3+ 2x2 − 6x − 3 = 0

Solution:

Hence root in interval [1,2]

Using Newton Raphson

Root = 1.93 to 2 decimal places.

f(0) = −3 f(x) = x3+ 2x2 − 6x − 3f(1) = 1 + 2 − 6 − 3 = −6f(2) = 8 + 8 − 12− 3 = 1

f(x) = x3+ 2x2 − 6x − 3

f ′(x) = 3x2 + 4x − 6x0 = 2

Then x1 = x0 −f(x0)

f ′(x0)

= 2 − 114

= 1.92857

x2 = 1.92857− 0.0404494

12.872427= 1.92857− 0.00314= 1.9254





Question:

Find the smallest positive root of the equation correct to two decimal places. Use the Newton–Raphson process.

x4 + x2 − 80 = 0

Solution:

Hence root = 2.91 to 2 decimal places.

f(x) = x4 + x2 − 80

f ′(x) = 4x3 + 2x

Let x0 = 3 f(3) = 10

So x1 = 3 −f(x0)

f ′(x0)

x1 = 3 − 10

114= 2.912

Then x2 = 2.912− 0.1768104.388

= 2.908





Question:

Apply the Newton–Raphson process to find the negative root of the equation correct to two decimal places. x3− 5x + 2 = 0

Solution:

Hence root between interval [−2,−3]

Let x0 = −2

Root = −2.44 correct to 2 decimal places.

f(x) = x3− 5x + 2

f ′(x) = 3x2 − 5

f(0) = 2f(−1) = −1 + 5 + 2 = 6f(−2) = −8 + 10+ 2 = 4f(−3) = −27+ 15+ 2 = −10

Then x1 = −2 −f(x0)

f ′(x0)

= −2 − 47

= −2.5714

x2 = −2.571− f(x1)

f ′(x1)

= −2.571− 2.1394

14.83= −2.4267

x3 = −2.4267− 0.157012.6662

= −2.4267− 0.01234= −2.439

x4 = −2.439− 0.0016312.846

= −2.4391





Question:

Show that the equation has a root in the interval [2, 3]. Taking 3 as a first approximation to this root, use the Newton–Raphson process to find this root correct to two decimal places.

2x3 − 4x2 − 1 = 0

Solution:

Sign change implies root in interval [2,3]

Let x0 = 3

Ans = 2.11 correct to 2 decimal place.

f(x) = 2x3 − 4x2 − 1.f(2) = 16− 16− 1 = −1f(3) = 54− 36− 1 = 17

f ′(x) = 6x2 − 8x

Then x1 = 3 −f(x0)

f ′(x0)

= 3 − 1730

= 2.43

x2 = 2.43− f(2.43)

f ′(2.43)

= 2.43− 4.078

16.05= 2.43− 0.254= 2.179

x3 = 2.179− f(2.179)

f ′(2.179)

= 2.179− 0.699811.056

= 2.179− 0.063296= 2.116

x4 = 2.116− f(2.116)

f ′(2.116)

= 2.116− 0.0388

9.937= 2.112

x5 = 2.112− f(2.112)

f ′(2.112)

= 2.112− −0.000849.8672

= 2.112





Question:

Taking 1.4 as a first approximation to a root, x, of this equation, use Newton–Raphson process once to obtain a second approximation to x. Give your answer to three decimal places.

f(x) = x3− 3x2 + 5x − 4

Solution:

Let x0 = 1.4


f(x) = x3− 3x2 + 5x − 4

f ′(x) = 3x2 − 6x + 5

x1 = 1.4− f(1.4)

f ′(1.4)

= 1.4− −0.136

2.48= 1.4+ 0.0548= 1.455 to 3 decimal places





Question:

Use the Newton–Raphson process twice to find the root of the equation which is near to . Give your answer to three decimal places.

2x3 + 5x = 70 x = 3

Solution:

Let x0 = 3


f(x) = 2x3 + 5x − 70

f ′(x) = 6x2 + 5

x1 = 3 − f(3)

f ′(3)

= 3 − −1

59= 3.02

x2 = 3.02− f(3.02)

f ′(3.02)

= 3.02− 0.1872

59.72−3.017 to 3 decimal places.



Solutionbank FP1 Edexcel AS and A Level Modular Mathematics Numerical solutions of equations Exercise D, Question 1


Question:

Given that has a root in the interval , use interval bisection on the interval to obtain the root correct to one decimal place.

f(x) = x3− 2x + 2 [−1 , − 2] [−1 , − 2]

Solution:

Hence root in interval [−1, −2] as sign change

Hence solution is −1.8 to 1 decimal place.

f(x) = x3− 2x + 2f(−1) = −1 + 2 + 2 = +3f(−2) = −8 + 4 + 2 = −2

a f(a) b f(b) a + b

2

f(a + b)

2

−1 +3 −2 −2 −1.5 +1.625−1.5 1.625 −2 −2 −1.75 0.141−1.75 0.141 −2 −2 −1.875 −0.842−1.75 0.141 −1.875 −0.841 −1.8125 −0.329−1.75 0.141 −1.8125 −0.329 −1.78125


3/18/2013file://C:\Users\Buba\kaz\ouba\fp1_2_d_1.html



Question:

Show that the equation has one positive and two negative roots. Obtain the positive root correct to three significant figures using the Newton–Raphson process.

x3− 12x − 7.2 = 0

Solution:

positive root between [3, 4]

negative roots between [0, −1], [−3, −4] Let x0 = 4

Using

where

So

Hence root = 3.73 to 3 significant figures

f(x) = x3− 12x − 7.2 = 0

f(0) = −7.2 f(−1) = 3.8f(1) = −18.2 f(−2) = 8.8f(2) = −23.2 f(−3) = 1.8f(3) = −16.2 f(−4) = −23.2f(4) = 8.8

x1 = x0 −f(x0)

f ′(x0)

f(x) = x3 − 12x − 7.2

f ′(x) = 3x2 − 12

x1 = 4 − 8.8

36

x1 = 3.756 to 3d.p.

x2 = 3.756− 0.716

30.322x2 = 3.732

x3 = 3.732− 0.01130.323

x3 = 3.7316





Question:

Find, correct to one decimal place, the real root of by using the Newton–Raphson process. x3+ 2x − 1 = 0

Solution:

Hence root interval [0, 1]

Using

Hence root is 0.5 to 1decimal place.

f(x) = x3+ 2x − 1f(0) = −1f(1) = 2

f(x) = x3+ 2x − 1

f ′(x) = 3x2 + 2 and x0 = 1

x1 = x0 −f(x0)

f ′(x0)

x1 = 1 − 2

5x1 = 0.6

x2 = 0.6− 0.4163.08

x2 = 0.465

x3 = 0.465− 0.0312.647

x3 = 0.453





Question:

Use the Newton–Raphson process to find the real root of the equation , taking as the first approximation and carrying out one iteration.

x3+ 2x2 + 4x − 6 = 0 x = 0.9

Solution:

f(x) = x3 + 2x2 + 4x − 6

f ′(x) = 3x2 + 4x + 4f(0.9) = −0.051

f ′(0.9) = 10.03

x1 = x0 −f(x0)

f ′(x1)

= 0.9− −0.051

10.03= 0.905 to 3 decimal places





Question:

Use linear interpolation to find the positive root of the equation correct to one decimal place. x3− 5x + 3 = 0

Solution:

Hence positive root in interval [1, 2] Using linear interpolation and x, as the 1st approximation

Then

root = 1.8 to 1 decimal place

f(x) = x3 − 5x + 3f(1) = −1f(2) = +1.

2 − x1x1− 1

= 11

2 − x1 = x1− 12x1 = 3

x1 = 1.5 f(x1) = 1.125

2 − x2x2 − 1.5

= 1

1.125

x2 = 1.882 f(x2) = 0.2601.882− x2

x2 − 1.5= 0.260

1.125

x2 = 1.810 f(x3) = −0.1171.882− x4x2 − 1.810

= 0.2600.117

= 1.832





Question:

.

a Show that the real root of lies in the interval [1, 2].

b Use the linear interpolation on the interval [1, 2] to find the first approximation to x.

c Use the Newton–Raphson process on f(x) once, starting with your answer to b, to find another approximation to x, giving your answer correct to two decimal places.

f(x) = x3+ x2 − 6

f(x) = 0

Solution:

a

Hence root in interval [1, 2]

b

c

f(x) = x3 + x2 − 6f(1) = −4f(2) = 6

2 − x1x1− 1

= 6

4

x1 = 1.4

x0 = 1.4

f(x) = x3+ x2 − 6

f ′(x) = 3x2 + 2x

x1 = x0 −f(x0)

f ′(x1)

= 1.4− −1.296

8.68= 1.55 to 2 decimal places





Question:

The equation has a root in the interval [1.0, 1.4]. Use linear interpolation once in the interval [1.0, 1.4] to find

an estimate of the root, giving your answer correct to two decimal places.

cosx = 1

4x

Solution:

cosx = 14

x ⇒ f(x) = 14

x − cosx

f(1) = −0.29f(1.4) = 0.1801.4− x1x1− 1

= −0.290

−0.180

x1 = 1.153

x1 = 1.15 to 2 decimal places





Question:

Use the Newton–Raphson process to find the positive root of this equation correct to two decimal places.

f(x) = x3− 3x − 6

Solution:

Hence root in interval [2, 3]

Let x0 = 2

Then

Ans = 2.28 to 2 decimal places

f(x) = x3− 3x − 6

f ′(x) = 3x2 − 3

f(0) = −5 f(1) = −7f(2) = −3 f(3) = +13

x1 = x0 −f(x0)

f ′(x1)

= 2 − −3

9x1 = 2.333

x2 = − 4.301

16.500x2 = 2.297

x3 = 2.297− 0.228

12.828x3 = 2.279

x4 = 2.279− −0.00023612.582

= 2.279+ 0.000019x4 = 2.2790



Documents

C:UsersBubakazouba p1 2 a 1 - solution banks sol FP1/CH2.… · Solutionbank FP1 Edexcel AS and A Level Modular Mathematics Numerical solutions of equations Exercise A, Question 2